如何在 Javascript 中合并和过滤两个对象数组

首先构建一个对象用于引用所有者，然后迭代汽车，并组合一个新对象。

var owners = [{ "ownerid": "11", "name": "jane", "sex": "female" }, { "ownerid": "22", "name": "mike", "sex": "male" }, { "ownerid": "33", "name": "alex", "sex": "male" }],
    cars = [{ "ownerid": "11", "make": "ford", "model": "mustang" }, { "ownerid": "11", "make": "honda", "model": "civic" }, { "ownerid": "33", "make": "toyota", "model": "corolla" }],
    obj = {},
    result;
owners.forEach(function (a) {
    obj[a.ownerid] = a;
});
result = cars.map(function (a) {
    return {
        name: obj[a.ownerid].name,
        sex: obj[a.ownerid].sex,
        make: a.make,
        model: a.model
    };
});
document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

var owners = [{"ownerid":"11", "name":"jane", "sex":"female"}, {"ownerid":"22", "name":"mike", "sex":"male"}, {"ownerid":"33", "name":"alex", "sex":"male"}];
var cars = [{"ownerid":"11", "make":"ford", "model":"mustang"}, {"ownerid":"11", "make":"honda", "model":"civic"}, {"ownerid":"33", "make":"toyota", "model":"corolla"} ];
var mergedandfiltered = [];
for(var i = 0; i < owners.length; i++){
  for(var j = 0; j < cars.length; j++){
    if(owners[i]['ownerid'] == cars[j]['ownerid']){
        mergedandfiltered.push({"name":owners[i]['name'],"sex":owners[i]['sex'],"make":cars[j]['make'],"model":cars[j]['model']});
    }        
  }  
}

为了提高效率，我们首先将用户列表转换为使用 owner 作为键的字典。我们可以使用 Array.protoype.reduce 来做到这一点：

var ownerDict = owners.reduce(function(p, c) {
  p[c.ownerid] = c;
  return p;
}, {});

这将为您提供一个如下所示的对象：

{
    "11": {
        "ownerid": "11",
        "name": "jane",
        "sex": "female"
    },
    "22": {
        "ownerid": "22",
        "name": "mike",
        "sex": "male"
    },
    "33": {
        "ownerid": "33",
        "name": "alex",
        "sex": "male"
    }
}

我们这样做的原因是，现在我们可以通过以下方式查找用户：

var owner = ownerDict[11];

这将返回对象：

{
    "ownerid": "11",
    "name": "jane",
    "sex": "female"
}

而那个抬头是O(1).如果你必须实际搜索数组才能找到它，假设它还没有排序，你的搜索将是O(n)（如果它被排序，你的示例数据实际上看起来是，它只是稍微好一点 - 你可以使用二叉树搜索，这是O(log(n))，但这仍然比O(1)慢，代码更复杂。

现在有了这个字典，我们现在可以使用 Array.prototype.map 将我们的cars数组转换为您想要的内容，如下所示：

var mergedandfiltered = cars.map(function(c) {
  var owner = ownerDict[c.ownerid];
  return {
    name: owner.name,
    sex: owner.sex,
    make: c.make,
    model: c.model
  };
});

var mergedandfiltered = [];
for (var i = 0; i < owners.length; i++) {
    var oID = owners[i].ownerid;
    for (var j = 0; j < cars.length; j++) {
      var mitem = {};
      cID = cars[j].ownerid;
      if (oID==cID) {
        mitem.name = owners[i].name;
        mitem.sex = owners[i].sex;
        mitem.make = cars[j].make;
        mitem.model = cars[j].model;
        mergedandfiltered.push(mitem);
      }
    }
}

var owners = [ {
        "ownerid" : "11",
    "name" : "jane",
    "sex" : "female"
}, {
    "ownerid" : "22",
    "name" : "mike",
    "sex" : "male"
}, {
    "ownerid" : "33",
    "name" : "alex",
    "sex" : "male"
} ];
var cars = [ {
    "ownerid" : "11",
    "make" : "ford",
    "model" : "mustang"
}, {
    "ownerid" : "11",
    "make" : "honda",
    "model" : "civic"
}, {
    "ownerid" : "33",
    "make" : "toyota",
    "model" : "corolla"
} ];
var merge = function(array, array2, filter) {
    var result = [];
    for ( var i in array) {
        for (var j in array2) {
            var filtered = filter(array[i],array2[j]);
            if (filtered != null){
                result.push(filtered);
            }
        }
    }
    return result;
};
var result = merge(owners,cars,function(owner,car){
    if (owner.ownerid == car.ownerid){
        return {
            name : owner.name,
            model : car.model
        };
    }
    return null;
});
console.log('array',result);

您可以将所有者数组转换为"哈希"（使用所有者作为键），然后根据 ownerid 从汽车数组中获取其属性。

var hashowners = [];
owners.forEach(function(o) {
   hashowners[o.ownerid] = {"name": o.name, "sex": o.sex};
});
var finalcars = cars.map( function(car) {
    car.name = hashowners[car.ownerid].name;
    car.sex = hashowners[car.ownerid].sex;
    delete car.ownerid;
    return car;
});
console.log(finalcars)

小提琴：https://jsfiddle.net/61fb13n8/